How the crap do u do this

A city bus collected $780 in fares on one day. The price of a regular fare was $0.80, and the price of a discount fare was $0.40. If a total of 1200 people paid the fares on this bus, how many people paid the regular fare?

I already know the answer is 750, but my question is, HOW THE CRAP DO I SOLVE THIS?
thanks :D

what are these kind of problems called btw, if you know

Regular fare (call it R) is .80
Discount fare (call it D) is .40
Number of riders is R + D = 1200
Total Fares earned is .80R + .40D = 780
Solve for either R or D
R +D - D = 1200 - D
R = 1200 - D
Substitue for R in the 2nd equation
.80(1200-D) + .40D = 780
960 - .80D + .40D = 780
960 - .40D = 780
960 - .40D + .40D - 780 = 780 + .40D -780
180 = .40D
180/.40 = .40D/.40D
450 = D
R + D = 1200
R + 450 = 1200
R + 450 - 450 = 1200 - 450
R = 750
PROOF
750(.80) + 450(.40) = 780
600 + 180 = 780
780 = 780

I think you can set a SYSTEM OF EQUATIONS with this.
assign the amount of regular fares as "x"
assign the amount of discount fares as "y"
since every 1 person in x pays $0.80 and ever 1 person in y pays $0.40, you can make the equation

0.80x + 0.40y = $780.00
x+y=1200 ( the amount of people paying a fare = 1200 people)

I'd use SUBSTITUTION here, but there is another way too. I'm gonna do substitution though.

y=1200-x (convert second equation into this)

.80x + .40(1200-x)=780 (substituted for y in the first equation)
.80x+ 480 - .40x=780
.40x= 300

x= 750

ask if your confused on any steps

algebra

r = number of regular fares
d = number of discount fares

r + d = 1200
d = 1200 -r
.8r + .4d = 780
.8r +.4(1200-r) = 780
.8r + 480 - .4r = 780
.4r = 300
r = 750
d = 1200-750 = 450